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A variation of this question has already been asked here, but I wish to solve via generating function.

My question's answer is equivalent to the bijection...

$$ card\left(\left\{\left[x_1\;x_2\;x_3\right]^T\in \mathbb{W}^3 : 1x_1 + 2x_2 + 3x_3 = 10\right\}\right) \\ = \left[x^{10}\right]\left(x^0 + x^1 + \cdots + x^{10}\right)\left(x^0 + x^2 + \cdots + x^{10}\right)\left(x^0 + x^3 + \cdots + x^9\right) $$

N. F. Taussig
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Landon
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  • Do you consider the order of the blocks to be significant? – awkward Mar 30 '19 at 12:36
  • @awkward yes i do. – Landon Mar 30 '19 at 16:29
  • If order is significant then you have the wrong bijection. The number of solutions to $x_1+2x_2+3x_3=10$ will give you the number of ways to construct the tower if you do not consider the order of the blocks to be significant. An exponential generating function might work better than an ordinary power series generating function in this case. – awkward Mar 30 '19 at 23:16
  • @awkward Ty for the feedback; I will start learning about exponential generating functions. Just for clarification, when you asked about ordering, it was a question of whether there should be a difference between, say, 3331 & 1333? & NOT a question of whether the ordering of the 3 consecutive 3s mattered? If so, then I replied correctly. – Landon Mar 31 '19 at 01:09
  • Yes, you have the correct understanding of ordering. 33 is the same as 33 (no distinction between 3's), but 13 is not the same as 31. – awkward Mar 31 '19 at 12:19

2 Answers2

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As no one's had a go at answering this I'll give it a shot. In the comments you state that the order of the blocks is important which makes me think you need to keep track of the different types of block which I don't think the set up of the answer you suggest in the question can do. So I think you need to consider, $$\left[x^{p}y^{q}z^{r}:p+q+r=10\right]$$$$\left(x^0 + x^1 + x^2 +\cdots \right)\left(y^0 + y^2 + y^4 +\cdots \right)\left(z^0 + z^3 +z^6+ \cdots \right)$$ Wolfram Alpha would expand this for you, but as there's only 14 bits, I'll give it a go by hand; $$x^{10}y^0z^0:1 : {10 \choose 1}$$ $$x^8y^2z^0:9: {9 \choose 1}or{9 \choose 8}$$ $$x^7y^0z^3:8:{8 \choose 1} or {8 \choose 7}$$ $$x^6y^4z^0:28:{8 \choose 2} or {8 \choose 6}$$ $$x^5y^2z^3:42: \frac{7!}{5!1!1!}$$ $$x^4y^6z^0:35: {7 \choose 3} or {7 \choose 4}$$ $$x^4y^0z^6:15: {6 \choose 2} or {6 \choose 4}$$ $$x^3y^4z^3: 60 : \frac{6!}{3!2!1!}$$ $$x^2y^8z^0:15: {6 \choose 2} or {6 \choose 4}$$ $$x^2y^2z^6:30: \frac{5!}{2!1!2!}$$ $$x^1y^6z^3:20:\frac{5!}{1!3!1!}$$ $$x^1y^0z^9:4 : {4 \choose 1}or{4 \choose 3}$$ $$x^0y^{10}z^0:1 : {10 \choose 0}$$ $$x^0y^4z^6: 6 : {4 \choose 2}$$ Phew!

I make that 274 different ways of building the tower.

Martin Hansen
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  • That's the answer I got when solving the recurrence relation...

    $$ a_n = a_{n - 1} + a_{n - 2} + a_{n - 3},;a_1 = 1,;a_2 = 2,;a_3 = 4 $$

    Thanks for your hard work!

     – Landon Mar 31 '19 at 00:54
  • I thought I was finally getting the hang of generating functions... Would you mind sharing any resources you found most helpful when learning about them? – Landon Mar 31 '19 at 00:59
  • OK : My answer both of these comments is too long for a comment; I'll add another answer.... – Martin Hansen Mar 31 '19 at 08:52
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Following on from your comment that you got the same answer by solving the recurrence relation, that too is amenable to a generating function approach.

Given you have that, $$a_n=a_{n-1}+a_{n-2}+a_{n-3}, a_1=1, a_2=2, a_3=4$$ Start with the deduction that $$a_{0}=1$$ then get the generating function (I can provide more detail if required), $$GF=\frac{1}{1-x-x^2-x^3}$$ which expands as, $$1+x+2x^2+4x^3+7x^4+13x^5+24x^6+44x^7+81x^8+149x^9+274x^{10}+504x^{11}+\dots$$ and there's the part we want just sitting there; $$274x^{10}$$ Resource : I expanded the brackets using the online Taylor Series Expansion Calculator at https://www.numberempire.com/taylorseriesexpansion.php

Other Generating Function Resources was the subject of a recent post on MSE at : How can I learn about generating functions?

Martin Hansen
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